# Proof: $\mathbb{R}^n$ is separable, show that every open cover of $\mathbb{R}^n$ has a countable subcover

Using the fact that $$\mathbb{R}^n$$ is separable, show that every open cover of $$\mathbb{R}^n$$ has a countable subcover.

I was given a hint to prove the following claim and if needed, incorporate it into the original proof: ''A set is open if and only if it is equal to an arbitrary collection of open balls", which I have been able to prove both ways. However, I don't see how I can use that to go about the original proof.

Otherwise, I thought of the following statements
$$(i)$$ $$\mathbb{R}^n$$ is separable
$$(ii)$$ $$\mathbb{R}^n$$ is compact
$$(iii)$$ Any open cover of $$\mathbb{R}^n$$ has a countable subcover

It would have been great if I could have proved $$(i)\Rightarrow(ii)\Rightarrow(iii)$$ but the problems that arise are
1. Firstly, $$\mathbb{R}^n$$ is not compact. Also, ''Every compact metric space is separable", which means $$(ii)\Rightarrow(i)$$.
2. ''For a compact space, every open cover has a finite subcover", which is not equivalent to countability.

How should I prove this? How are separability and claimed fact requirements to the proof?

Edit. The attempted second method is obviously wrong because $$\mathbb{R}^n$$ is not compact.

Cover $$\Bbb R^n$$ with any countable collection of compact sets (e.g. all closed balls of radius $$\sqrt n$$ centred on points with integral coordinates). Now just cover each of these sets separately with a finite subcover. A countable union of finite covers is countable.

I don't see where I've used the separability of $$\Bbb R^n$$, but I'm sure it's hidden somewhere in there...

• Compact subsets of \Bbb R^n are separable so a countable union of compact subsets of \Bbb R^n is separable. Dec 3, 2020 at 19:59
• I think it's fair to say you haven't used the separability of $\mathbb{R}^n$. Rather what you've done is show that every $\sigma$-compact space is has the "Lindelöf property" (every open cover admits a countable subcover). Even a compact space doesn't need to be separable. Dec 3, 2020 at 20:32

$$\mathbb R^n = \bigcup_{x \in \mathbb Z^n} ([0,1]^n + x)$$, write $$A_x = [0,1]^n + x$$. The $$A_x$$ cover $$\mathbb R^n$$. Let $$\mathcal U$$ be a cover of $$\mathbb R^n$$, then $$\mathcal U$$ covers $$A_x$$ and has a finite subcover $$\mathcal U_x$$. The set $$\bigcup_{x \in \mathbb Z^n} \mathcal U_x$$ is a countable cover.

Here are generalizations, as mentioned in a comment:

1. Every hemicompact space is countably compact.
2. Every separable metric space is countably compact.
• Separable metrizable spaces are 2nd-countable, and 2nd-countable spaces are Lindelof. Dec 3, 2020 at 20:01